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#1

# Does the voltage divider in feedback loop contribute to loop gain?

I am confused by the voltage divider ratio used in loop gain calculation. Some textbook and application note include this ratio in the open loop gain, but some does not. I was convinced that the the ratio should not be considered in analog control, however, when I design a digital controller for SMPS, coefficient of proportional control is related to the ratio. So, my question is should I include the voltage divider ratio in loop gain? Is it different from analog control to digital control?
08-23-2013 09:38 PM
Top #2
It is not clear what you are trying to do. A voltage divider in the feedback loop does affect the closed loop gain. However, open loop gain is with the feedback "not closed". This means for the open loop analysis, the output of the voltage divider would not go anywhere, and would thus, have no affect except for the load of the voltage divider on the circuit to which it is attached.

This, highlights the need to understand and analyze a circuit, not just find values to plug into formulas.
08-23-2013 09:39 PM
Top #3
This depends "where" the divider is placed.
In different text-books it is done differently...
If the lower divider resistor is placed within the "Virtual Ground" of the Error-Amp, then it only affects the DC set-point and does not effect the AC gain, within reason...
If the sampling divider is placed PRIOR to the input resistor of the Error-Amp, then you would have a AC gain loss to include in your closed loop analysis.... You will see this in the Pressman book ... I prefer not to do the later, since you will need to make up this GAIN somewhere else and that would put more burden on the Error-Amp..
There are cases where it may be needed...but typically not....
08-23-2013 09:39 PM
Top #4
In the classic application of the divider in a feedback loop with an op amp, the divider ratio has no effect on the gain of the loop. There are quite a few publications and books out there that get this wrong, including some by some prominent authors.

Perhaps you can post your specific circuit.
08-23-2013 09:40 PM
Top #5
I am sorry I did not make my question clear. The voltage divider is used at the output of converter parallel with load to compare with vref of control chip. I don't know whether I get it right from your comment. The voltage divider has effect on close loop gain but no effect on open loop gain, am I right? If so, for using the bode plot to analyze stability(open loop), I should not include voltage divider ratio.
08-23-2013 09:41 PM
Top #6
Excellent points, and I stand corrected. I sometimes use op-amps with a divider on the non-inverting input in loops (not necessarily power supply circuits) where the ratio does matter, but the examples I gave were circuits with virtual grounds where it does not. Siyu, I apologize for the confusion.
08-23-2013 09:41 PM
Top #7
Hi, John
I read through the application note, but I don't seen the lower divider resistor Rf2 in Type II and Type III shown in transfer function. To my understanding, the voltage divider composed of Rf1 and Rf2 does not have AC gain in the loop. Am I right?
08-23-2013 09:42 PM
Top #8
Hi, Chris
I read Basso's book, in which he showed the whole process why the lower divider resistor should be ignored in AC analysis. However, I just tried to control a converter by DSP. Take just proportional control for instance to explain my confusion, the sample output voltage is 52 Volts. Method 1: vref=2.5, voltage divider ratio=1/20. voltage error= 52/20-2.5=0.1. Method 2: vref(in dsp)=50, voltage error=52-50=2. If I need to get same control signal form boht methods, I need to take divider ration into consideration. So, I am thinking about the voltage divider might have effect on loop gain.

Hi Ray,
Thanks for your reply. I am designing digital control for converter by using DSP. The circuit is just a basic boost converter. In digital control, there is no op-amp to work with voltage divider. Would this change the conclusion?
08-23-2013 09:43 PM
Top #9
It would be easier if you divided you whole loop into separate gain blocks of transfer functions such as: PWM Gain, Output filter, Vsense & EA, etc, etc.
For your Vsense & EA block, write an equation of the transfer function from the input of the first resistor, R1 (at the Vout sense) to the output of the error amp. If the resistor divider (R1 and R2) goes directly into the -input of the error amp, and the +input of the error amp is your Vref signal, your equation will show you that R2 is not part of the AC transfer function.
For many applications though, you may find that you need to step up to a type III comopensation or, you may need to filter noise on the voltage divider before sending it to the error amp. In these cases you still want to write an equation for that gain block.
08-23-2013 09:43 PM
Top #10
Yes it would change things. When the divider feeds an opamp, the midpoint of the divider is a virtual ground. Hence the lower resistor is a short and not part of the gain equation.

If you are just sampling the midpoint with your DSP, the attenuation will become part of the loop equation.

When you are done with it all, its a good idea to measure the loop in the old fashioned analog way anyway. Remember, power supplies and controllers often don't behave as they should according to the theory.

The measurement will confirm, of course, whether the divider is doing what you think it should.
08-23-2013 09:44 PM
Top #11
OPEN LOOP RESPONSE also known as PLANT RESPONSE , in your case does not include the sample network, which is the voltage divider.
The voltage divider is included in the FEEDBACK NETWORK......
STABILITY ANALYSIS looks at the entire CLOSED LOOP SYSTEM.....this is the FEEDBACK NETWORK and the PLATE RESPONSE....
Depending on where you place your divider, it will or will not affect the AC gain of the FEEDBACK...
Could you provide example of how your divider and error amplifier is configured...
08-23-2013 09:44 PM
Top #12
Hi Darrel,
I used the method as you said and I was convinced by transfer function in which the lower divider resistor is not shown. However, when I came to design a digital controller and looked at the discrete value used in compensation caculation, I noticed that it might different with analog control. Take just proportional control for instance to explain my confusion, the sample output voltage is 52 Volts. Method 1: vref=2.5, voltage divider ratio=1/20. voltage error= 52/20-2.5=0.1. Method 2: vref(in dsp)=50, voltage error=52-50=2. If I need to get same control signal form boht methods, I need to take divider ration into consideration
08-23-2013 09:45 PM
Top #13
To answer the specific question that you asked me, you are correct that in the app note the voltage divider ratio does not figure into the AC analysis for the reason that others have pointed out. I should have looked at the analysis more closely, but in my haste, I did not. I hope that reading through the app note provided some benefit and it did not waste your time. In any event, I apologize for the contradiction in my post and any confusion that it may have caused.
08-23-2013 09:46 PM
Top #14
This is a good discussion. Similar in my case also. I tried to analysis closed loop transfer function of SMPS which is implementing with digital controller. No error amplifier or any type of compensator circuit implemented here.So in that case, we can find/measure the open loop transfer in the analog way. But how to get on closed loop analysis with DSP ?? We will be helpful If someone can easily explain or share some good documents.
08-23-2013 09:46 PM
Top #15
Hi Chris,
I was trying to plot frequency response of overall loop(plant and feedback). In my DSP implementation, the voltage divider is paralled with load, let's say R1 is the upper resistor and R2 is the lower one. DSP is sampling at the connection between R1 and R2, in this way the output voltage would fall in to the range of 0-3V.
I don't know how to upload image of my circuit. Does linkedin has this function?

With AP 102B, I am still thinking about a way to measure response from input of feedback(or load) to the output of voltage compensator.
P.S. I am using current mode control.
08-23-2013 09:47 PM
Top #16
My 2 cents.

When a voltage error amplifier is used for the voltage feedback loop control, I agree with Ray as well as the reference to Chris Basso's book (amongst others, but I'm most familiar with Basso's book) "Switch-Mode Power Supplies-SPICE Simulations and Practical Designs" that the resistor divider does not affect the closed loop response. Chris Basso does nice work describing the virtual ground.

HOWEVER, if an OTA (operational transconductance amplifier) is used, the voltage divider MUST be included in the loop response. I believe Chris Basso also describes this in his book.

From a DSP aspect, I'm only guessing it sounds to me like the digital response will be more aligned with a voltage error amplifer type of response. If the DSP app notes don't cover the topic, perhaps a call to the local FAE could provide you a fast track to an expert who would know.
08-29-2013 09:56 PM
Top #17
OK... I see...
Controller with ADC input... If you feed the ADC straight from the divider, then yes the divider needs to be accounted for, since this is Gain loss...
But keep in mind that the ADC will need some protection against over voltage and under voltage events that will make the SAR go poof !!!!
I typically use a Unity gain follower with the Vdd tied to the same reference voltage as the ADC.... and the Vss to ground...assuming your ADC input is at ground reference...
This will limit swing to a safe region, just in case...
08-29-2013 09:57 PM
Top #18
It sounds like you know how to measure your DSP-controlled SMPS, but are not sure how to calculate the closed-loop response. This is not something that can be explained in this forum or even in a several-page app-note. You would need a course in complex variables, then a course or courses in discrete-time signal processing, which are both involved topics. One problem is that continuous-time (S-domain) goes from DC (0) to infinity in frequency, while in the discrete-time (Z-domain) it goes from 0 to pi (or -pi to +pi, where the 0 to -pi side is a duplicate of the 0 to +pi side). In this case, pi represents the Nyquist frequency, or 1/2 the sampling frequency. There is a BiLinear Transform to convert from the S to the Z domains, but the specific poles and zeros, especially at the higher frequencies need to be "pre-warped' so that they map as correctly as possible from the S to Z domains or vice-versa. This assumes that the sampling frequency is high-enough to do this. Beyond that, the math in the Z-domain is somewhat different than the S-domain. At the very least, I'd recommend that you get a good book like Oppenheim & Schafer's "Discrete-Time Signal Processing".